
\begin{center}
\begin{tikzpicture}[scale=.9,every node/.style={scale=0.9}]
\coordinate (A) at (0,0);
\coordinate (B) at (4,0);
\coordinate (C) at (1,2);
\coordinate (D) at (-3,2);
\coordinate (I) at (barycentric cs:A=1,C=1);
\coordinate (M) at ($(A)+(D)-(I)$);
\coordinate (N) at ($(B)+(C)-(I)$);
\draw (A) node[below] {A} -- (B) node[below] {B} -- (C) node[above] {C} -- (D) node[above] {D} -- cycle;
\draw (A) -- (C);
\draw (D) -- (B);
\fill (I) circle (2pt) node[below right] {I};
\fill (M) circle (2pt) node[above] {M};
\fill (N) circle (2pt) node[above] {N};
\draw[->,>=stealth',red] (A) -- (M);
\draw[->,>=stealth',blue] (B) -- (N);
\end{tikzpicture}
\end{center}

\medskip

\begin{enumerate}
\setcounter{enumi}{1}
\item $\vv{IM}+\vv{IN}=\vv{IA}+\vv{IB}+\vv{IC}+\vv{ID}$.\\
Or, $\vv{ID}=-\vv{IB}$ et $\vv{IC}=-\vv{IA}$. \\
Donc, \underline{$\vv{IM}+\vv{IN}=\vv{0}$}.

\medskip

On en déduit alors que I est le milieu de [MN].

\item On sait que $\vv{IN}=\vv{IB}+\vv{IC}$ (par construction) et que $\vv{IN}=\vv{IB}+\vv{BN}$ (d'après la relation de Chasles).\\
Ainsi, $\vv{IB}+\vv{BN}=\vv{IB}+\vv{IC}$ et donc $\vv{BN}=\vv{IC}$.

\medskip

I étant le milieu de [AC], $\vv{IC}=\vv{AI}$.

\medskip

Ainsi, $\vv{IC}=\vv{AI}=\vv{BN}$ donc \underline{ABNI est un parallélogramme}.
\end{enumerate}